The very first thing is to division the triangle right into vertical strips of width dx and mass dm. Let"s think about one strip, whose center is in ~ a street x-from the origin.

You are watching: An 800 g steel plate has the shape of the isosceles triangle shown in the figure(figure 1).


Area that the triangle is provided by:

A = (1/2)bh = (1/2)(20 × 10-2)(30 × 102) = 0.03 m2

Let the size of the tiny strip be l.

The area of this piece is to express as:

dA = l·dx

The mass of the strip/the facet is:

dm = (M/A)dA = (800 × 10-3/0.03)ldx = 26.67 l·dx kg/m2

From proteries of similar triangles:

l/20 = x/30

l = (20/30)x = (2/3)x

Substituting to the expression that dm:

dm = (26.67)(2/3)x dx

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Problem Details

An 800 g steel plate has the form of the isosceles triangle displayed in number P12.53. What room the x-and y- works with of the center of mass?


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Based on ours data, us think this difficulty is pertinent for Professor Murray's class at GT.

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